A recent question on the notion and notation of multiplicative integrals
( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:

Now the question: How to evaluate the above integral
by means of standard integral
calculus?

I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.

Pietro, are you serious about "it would be a remarkable case of an integral that one can only evaluate directly from the definition of Riemann sums"? You just need a right representation of the logarithm of gamma function... (My turn for going to bed.) Have you tried Mathematica or Maple (symbolically)?
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Wadim ZudilinJul 22 '10 at 14:35

I undersand what you mean (I didn't mean to make a strong statement). Actually, following your suggestion, if we integrate on $[0,1]$ by series with this representation: $$\log(x!)=-\gamma\, x + \sum_{k=1}^{\infty}\left(\frac{x}{k}-\log\big(1+ \frac{x}{k}\big)\right)$$ and use the Stirling formula we find again $\frac{1}{2}\log(2\pi)-1.$
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Pietro MajerJul 23 '10 at 21:38

So we have three elementary integrals to deal with, plus
$$\int_0^{1/2} \log \sin(\pi z) \ dz. \quad (*)$$
According to Mathematica, $(*) = - \log(2)/2$. So, if we can find a clean proof of this fact, we will have evaluated the integral. This may be difficult, because the indefinite integral $\int \log \sin(\pi z) \ dz$ involves dilogarithms. To me, $(*)$ looks like a good target for residues. Anyone want to finish it off?

Nice to see an elegant reduction of the latter "non-elementary" integral. I worked it in on my way to the office in a more elementary way: after the change $t=\sqrt{\sin(\pi x)}$ the integral (up to constant) becomes $F'(y)|_{y=0}$ where $$F(y)=\int_0^1t^{1/2+y/2}(1-t)^{1/2}dt,$$ the Euler beta integral.
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Wadim ZudilinJul 22 '10 at 23:28